把1,8,9,13,14,22,26,28这8个数每6个组成一组的组法是什么 把{1,2,3,4,5,6,7,8,9}这九个数每三个一组,...
\u6025\u6c42C\u8bed\u8a00\u7a0b\u5e8f\uff1a\u5c061,2,3,4,5,6,7,8,9\u51719\u4e2a\u6570\u5206\u6210\u4e09\u7ec4,\u7ec4\u62103\u4e2a\u4e09\u4f4d\u6570,\u4e14\u4f7f\u8fd93\u4e2a\u4e09int check(int a,int b,int c){ int test[9]={0},i; long num = a*1000000+b*1000+c; if(b>1000 || c>1000)return 0;//\u786e\u4fdd\u662f\u4e09\u4f4d\u6570 for(i = 0;i < 10;i++) { int temp = num%10; num/=10; temp--; if(test[temp] == 0) { test[temp] = 1; } else return 0; } return 1;}int main(){ int i,j,k; int num; for(i = 1;i < 10;i++) { for(j =1;j < 10;j++) { if(i!=j) { for(k = 1; k < 10;k++) { if(k!=j) { num = i+j*10+k*100; if(check(num,num*2,num*3)){ printf("%d %d %d\n",num,num*2,num*3); } } } } } } return 0;}\u4e0a\u9762\u521a\u5199\u7684\uff0c\u5b9e\u73b0\u601d\u8def\uff1a\u5148\u628a\u6240\u6709\u4e09\u4f4d\u6570\u627e\u51fa\u6765\uff08\u5e76\u4e14\u4e09\u4f4d\u6570\u4e0d\u80fd\u76f8\u540c\uff09\uff0c\u4e0a\u9762\u4ee3\u7801\u7684\u5173\u952e\u5728check\u65b9\u6cd5\uff0c\u6211\u5148\u628a\u5b83\u4f20\u9012\u8fdb\u6765\u7684\u4e09\u4e2a\u6570\u636e\u8fdb\u884c\u68c0\u67e5\uff0c\u56e0\u4e3a\u4f20\u9012\u8fc7\u6765\u7684\u65f6\u5019\u5c31\u662f\u6784\u62101:2:3\u7684\u6bd4\u4f8b\uff0c\u6211\u73b0\u5728\u5c31\u662f\u8981\u68c0\u67e5\u662f\u5426\u662f\u4ece1\u52309\u6240\u6709\u6570\u5b57\u90fd\u6709\u5e76\u4e14\u4e0d\u91cd\u590d\uff08\u4e0d\u61c2\u5c31\u8ffd\u95ee\u3002\uff09
#include #include int main() { using namespace std; int arr[9] = {1, 2, 3, 4, 5, 6, 7, 8, 9}; sort(arr, arr+9); do // \u5168\u6392\u52171-9\uff0c\u5fc5\u987b\u4fdd\u8bc1arr\u4e3a\u5347\u5e8f { // \u524d3\u4e2a\u5143\u7d20\u7ec4\u6210a\uff0c\u4ee5\u6b64\u7c7b\u63a8 int a = arr[0]*100 + arr[1]*10 + arr[2]; int b = arr[3]*100 + arr[4]*10 + arr[5]; int c = arr[6]*100 + arr[7]*10 + arr[8]; if ( a+b == c) { cout << a << " + " << b << " = " << c << endl; } } while ( next_permutation(arr, arr+9) ); // \u53d6\u5f97\u4e0b\u4e00\u4e2a\u6392\u5217 return 0;}
可以组成28组。
共8个数,组成6个一组可组成计算式:
8*7*6*5*4*3/1*2*3*4*5*6=28组
可换成简单计算式:
8*7/1*2=28组
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